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Say, we want to remove discontinuity of $$\frac{x}{e^x-1}$$

As far as I know, a discontinuity of $f$ at the point $h$ (for $h$ole) is removable iff

$$\lim_{x\to h+} f(x) = \lim_{x\to h-} f(x) \neq \pm \infty$$

So I was wondering if it would be enough to write the following?

\begin{align} \lim_{x\to 0} \frac{x}{e^x-1} &= \Big[\frac{0}{0}\Big] \\ &= \lim_{x\to 0} \frac{1}{e^x} \\ &= 1 \end{align}

Should we ever care about $\lim_{x\to 0+}$ and $\lim_{x\to 0-}$ while applying l'Hôpital's rule?

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In this very case it is even simpler: the limit (not one sided!) exists, so you don't even need to split the calculation in two steps!

And yes: apply l'Hospital directly to the limit.

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